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davidmckenna

  • 4 years ago

Simplify x^2 - 9 / 2x^2 - 3x -9

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  1. davidmckenna
    • 4 years ago
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    Sorry I'm trying to write this out in proper equation

  2. saifoo.khan
    • 4 years ago
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    \[\frac{ x^2 - 9}{2x^2 - 3x -9}\]

  3. saifoo.khan
    • 4 years ago
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    Like that right?

  4. davidmckenna
    • 4 years ago
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    Yes that's it

  5. davidmckenna
    • 4 years ago
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    Couldn't dot he divide thing

  6. saifoo.khan
    • 4 years ago
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    \[\frac{(x-3)(x+3)}{(x-3) (2 x+3)}\] \[\frac{\cancel{(x-3)}(x+3)}{\cancel{(x-3)} (2 x+3)}\]

  7. saifoo.khan
    • 4 years ago
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    \[\frac{x+3}{2x+3}\]

  8. davidmckenna
    • 4 years ago
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    Could you explain like step by step, sorry I'm not very good at this

  9. saifoo.khan
    • 4 years ago
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    What u have to do is, u have to factor the both numerator and denominator..

  10. saifoo.khan
    • 4 years ago
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    After factoring, u will get, \[\frac{(x-3)(x+3)}{(x-3) (2 x+3)}\] Now u have to cancel the like terms, leaving u with, \[\frac{x+3}{2x+3}\]

  11. davidmckenna
    • 4 years ago
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    Thankyou I get it now, as soon as you factorised it's easy from there thanks a lot!

  12. saifoo.khan
    • 4 years ago
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    No problem! ^_^

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